If the rate of reaction is `2.6xx10^-3 mol L^-1s^-1` at `50^@C` and `7.02xx10^-2 mol L^-1s^-1` at `80^@C`: then what will be the temperature coefficient of the reaction ?

Explanation

`k_2/k_1 = (mu)^((80-50)/(10)) implies (7.02xx10^-2)/(2.6xx10^-3) = (mu)^3`
`implies 27 =(mu)^3 implies 3^3 = (mu)^3 = mu =3`